Kinetic Theory of Gases Equation (Edexcel A Level Physics)

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Lindsay Gilmour

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Kinetic Theory of Gases Equation

  • Gases consist of atoms or molecules randomly moving around at high speeds
    • The kinetic theory of gases models the thermodynamic behaviour of gases by linking the microscopic properties of particles (mass and speed) to macroscopic properties of particles (pressure and volume)

 

  • The theory is based on the following assumptions:
    • Molecules of gas behave as identical, hard, perfectly elastic spheres
    • The volume of the molecules is negligible compared to the volume of the container
    • The time of a collision is negligible compared to the time between collisions
    • There are no forces of attraction or repulsion between the molecules
    • The molecules are in continuous random motion

  • The number of molecules of gas in a container is very large, therefore the average behaviour (eg. speed) is usually considered

Root-Mean-Square Speed

  • The pressure of an ideal gas equation includes the mean square speed of the particles:

<c2>

  • Where
    • c = average speed of the gas particles
    • <c2> has the units m2 s-2

  • Since particles travel in all directions in 3D space and velocity is a vector, some particles will have a negative direction and others a positive direction
    • When there are a large number of particles, the total positive and negative velocity values will cancel out, giving a net zero value overall
    • In order to find the pressure of the gas, the velocities must be squared meaning that all the values end up positive

  • To calculate the average speed of the particles in a gas, take the square root of the mean square speed:

square root of less than c squared greater than end root equals c subscript r m s end subscript

  • cr.m.s is known as the root-mean-square speed and still has the units of m s-1

Worked example

An ideal gas has a density of 4.5 kg m-3 at a pressure of 9.3 × 105 Pa and a temperature of 504 K.

Determine the root-mean-square (r.m.s.) speed of the gas atoms at 504 K.

Step 1:            Write out the equation for the pressure of an ideal gas with density

p equals 1 third rho less than c squared greater than

Step 2:            Rearrange for mean square speed

less than c squared greater than equals fraction numerator 3 p over denominator rho end fraction

Step 3:            Substitute in values

less than c squared greater than equals fraction numerator 3 space cross times space left parenthesis 9.3 space cross times space 10 to the power of 5 right parenthesis over denominator 4.5 end fraction space equals space 6.2 space cross times space 10 to the power of 5 space m squared s to the power of negative 2 end exponent

Step 4:            To find the r.m.s value, take the square root of the mean square speed

c subscript r m s end subscript space equals space square root of less than c squared greater than end root space equals space square root of 6.2 cross times 10 to the power of 5 space m to the power of 2 space end exponent s to the power of negative 2 end exponent end root space equals space 787.4 space m space s to the power of negative 1 end exponent

Step 5:            Write the answer to the correct significant figures and include units

crms = 790 ms−1 (2 sig figs)

Deriving the Equation for Kinetic Theory

  • When molecules rebound from a wall in a container, the change in momentum gives rise to a force exerted by the particle on the wall
    • Many molecules moving in random motion exert forces on the walls which create an average overall pressure, since pressure is the force per unit area

The Situation for the Derivation

  • Picture a single molecule in a cube-shaped box with sides of equal length l
  • The molecule has a mass m and moves with speed c, parallel to one side of the box
  • It collides at regular intervals with the ends of the box, exerting a force and contributing to the pressure of the gas
  • By calculating the pressure this one molecule exerts on one end of the box, the total pressure produced by all the molecules can be deduced

Single molecule in box

A single molecule in a box collides with the walls and exerts a pressure

Five-Step Derivation

Step 1: Find the change in momentum as a single molecule hits a wall perpendicularly

  • One assumption of the kinetic theory is that molecules rebound elastically
    • This means there is no kinetic energy lost in the collision
    • If they rebound in the opposite direction to their initial velocity, their final velocity is -c
    • The change in momentum is therefore:

capital delta p space equals space minus m c space minus space left parenthesis plus m c right parenthesis space equals space minus m c space minus space m c space equals space minus 2 m c

Step 2: Calculate the number of collisions per second by the molecule on a wall

  • The time between collisions of the molecule travelling to one wall and back is calculated by travelling a distance of 2l with speed c:

t i m e space b e t w e e n space c o l l i s i o n s space equals space fraction numerator d i s t a n c e over denominator s p e e d end fraction equals space fraction numerator 2 l over denominator c end fraction

  • Note: c is not taken as the speed of light in this scenario

Step 3: Find the change in momentum per second

  • The force the molecule exerts on one wall is found using Newton’s second law of motion:

F o r c e space equals space r a t e space o f space c h a n g e space o f space m o m e n t u m space equals space fraction numerator capital delta p over denominator capital delta t end fraction space equals space fraction numerator 2 m c over denominator begin display style bevelled fraction numerator 2 l over denominator c end fraction end style end fraction space equals space fraction numerator m c squared over denominator l end fraction

  • The change in momentum is +2mc since the force on the molecule from the wall is in the opposite direction to its change in momentum

Step 4: Calculate the total pressure from N molecules

  • The area of one wall is l2
  • The pressure is defined using the force and area:

p r e s s u r e comma space p space equals space fraction numerator f o r c e over denominator a r e a end fraction space equals space fraction numerator begin display style bevelled fraction numerator m c squared over denominator l end fraction end style over denominator l squared end fraction space equals space fraction numerator m c squared over denominator l cubed end fraction

  • This is the pressure exerted from one molecule
  • To account for the large number of N molecules, the pressure can now be written as:

p space equals space fraction numerator N m c squared over denominator l cubed end fraction

  • Each molecule has a different velocity and they all contribute to the pressure
  • The mean squared speed of c2 is written with left and right-angled brackets <c2>
  • The pressure is now defined as:

p space equals space fraction numerator N m less than c squared greater than over denominator l cubed end fraction

Step 5: Consider the effect of the molecule moving in 3D space

  • The pressure equation still assumes all the molecules are travelling in the same direction and colliding with the same pair of opposite faces of the cube
  • In reality, all molecules will be moving in three dimensions equally
  • Splitting the velocity into its components cx, cy and cz to denote the amount in the x, y and z directions, c2 can be defined using pythagoras’ theorem in 3D:

c squared space equals space c x squared space plus space c y squared space plus space c z squared

  • Since there is nothing special about any particular direction, it can be determined that:

less than c subscript x squared greater than space equals space less than c subscript y squared greater than space equals space less than c subscript z squared greater than

  • Therefore, <cx2> can be defined as:

less than c subscript x squared greater than space equals space 1 third less than c squared greater than

  • The box is a cube and all the sides are of length l
    • This means l3 is equal to the volume of the cube, V

  • Substituting the new values for <c2> and l3 back into the pressure equation obtains the final equation:

p V space equals space 1 third N m less than c squared greater than

  • This is known as the Kinetic Theory of Gases equation:

Pressure of an ideal gas equation

  • It can also be written using the density ρ of the gas:

rho space equals space fraction numerator m a s s over denominator v o l u m e end fraction space equals space fraction numerator N m over denominator V end fraction

  • Rearranging the pressure equation for p and substituting the density ρ:

p space equals space 1 third rho less than c squared greater than

Examiner Tip

Make sure to revise and understand each step for the whole of the derivation, as you may be asked to derive all, or part, of the equation in an exam question.

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Lindsay Gilmour

Author: Lindsay Gilmour

Expertise: Physics

Lindsay graduated with First Class Honours from the University of Greenwich and earned her Science Communication MSc at Imperial College London. Now with many years’ experience as a Head of Physics and Examiner for A Level and IGCSE Physics (and Biology!), her love of communicating, educating and Physics has brought her to Save My Exams where she hopes to help as many students as possible on their next steps.